Respuesta :
Proof 1: Proving the composition of injective functions is itself an injective function.
f(x) is injective
f(a) = f(b) <--> a = b
g(x) is injective
g(a) = g(b) <--> a = b
prove f(g(x)) is injective
f(g(a)) = f(g(b))
f(a) = f(b) ... since g(x) is injective
a = b
So that proves if f(g(a)) = f(g(b)), then a = b
Now go the other direction
a = b
g(a) = g(b)
f(g(a)) = f(g(b))
So that wraps up the sub-proof of "f(g(a)) = f(g(b)) iff a = b"
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proving g(f(x)) is injective is going to follow a similar path of steps
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Proof 2: Proving the composition of surjective functions is itself a surjective function.
f(x) is surjective
The range of f(x) is the set of all real numbers
g(x) is surjective
The range of g(x) is the set of all real numbers
f(g(x)) is also surjective because of property 1 listed above. The range of f(x) is the set of all real numbers. The fact that g(x) represents all reals as well means that we touch all of the inputs needed to generate all the outputs for f(x).
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Proving g(f(x)) to be surjective follows a similar path.
f(x) is injective
f(a) = f(b) <--> a = b
g(x) is injective
g(a) = g(b) <--> a = b
prove f(g(x)) is injective
f(g(a)) = f(g(b))
f(a) = f(b) ... since g(x) is injective
a = b
So that proves if f(g(a)) = f(g(b)), then a = b
Now go the other direction
a = b
g(a) = g(b)
f(g(a)) = f(g(b))
So that wraps up the sub-proof of "f(g(a)) = f(g(b)) iff a = b"
-------------------------
proving g(f(x)) is injective is going to follow a similar path of steps
------------------------------------------
------------------------------------------
Proof 2: Proving the composition of surjective functions is itself a surjective function.
f(x) is surjective
The range of f(x) is the set of all real numbers
g(x) is surjective
The range of g(x) is the set of all real numbers
f(g(x)) is also surjective because of property 1 listed above. The range of f(x) is the set of all real numbers. The fact that g(x) represents all reals as well means that we touch all of the inputs needed to generate all the outputs for f(x).
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Proving g(f(x)) to be surjective follows a similar path.
